Page images

Hence cot A : cot B : : tang B : tang A ; that is, the colangents of two arcs are reciprocally proportional to their tangents.

The formula cot A x tang A=R* might be deduced immediately, by comparing the similar triangles CAT, CDS, which give AT : CA : : CD : DS, or tang A :R :: R : cot A

XIX. The sines and cosines of two arcs, a and b, being given, it is required to find the sine and cosine of the sum or difference of these arcs. Let the radius AC=R, the arc


D AB=a, the arc BD=b, and con

B sequently ABD=a + b. From

L the points B and D, let fall the

M perpendiculars BE, DF upon AC; L from the point D, draw DI perpendicular to BC ; lastly, from the point I draw IK perpendicu

A lar, and IL parallel to, AC. F CFK' KE P The similar triangles BCE, ICK, give the proportions,

sin a cos b. CB : CI : : BE: IK, or R: cos b :: sin a : IK=


cos a cos b. CB : CI :: CE : CK, or R: cos b :: cos a: CK=

R The triangles DIL, CBE, having their sides perpendicular, each to each, are similar, and give the proportions, CB : DI :: CE : DL, or R: sin b :: cos a : DL=


sin a sin b. CB : DI : : BE : IL, or R: sin b :: sin a : IL=

Ꭱ . But we have

IK+DL=DF=sin (a+b), and CK-IL=CF=cos (a+b). Hence

sin a cos b+sin b cos a
sin (a+b) =

cos a cos bsin a sin b.
cos (a+b)=

R The values of sin (ab) and of cos (246) might be easily deduced from these two formulas; but they may be found directly by the same figure. For, produce the sine DI till it meets the circumference at M; then we have BM=BD=b, and MI=ID=sin b. Through the point M, draw MP perpendicular, and MN parallel to, AC: since MI=DI, we have MN =IL, and IN-DL. But we have IK_IN=MP=sin: (amb), and CK+MN=CP=cos (ab); hence

cos a sin b.


[ocr errors]


[ocr errors]
[ocr errors]


sin a cos b-sin b cos a
sin (


cos a cos b+sin a sin b. cos (ab)=

R These are the formulas which it was required to find. The preceding demonstration may seem defective in point of generality, since, in the figure which we have followed, the arcs a and b, and even a+b, are supposed to be less than 90°. But first the demonstration is easily extended to the case in which a and b being less than 90°, their sum a+b is greater than 90°. Then the point F would fall on the prolongation of AC, and the only change required in the demonstration would be that of taking cos (a+b)

but as we should, at the same time, have CF=I'L'—CK', it would still follow that cos (a+b)=CK'—I'L', or R cos (a+b)=cos a cos b—sin a sin b. And whatever be the values of the arcs a and b, it is easily shown that the formulas are true: hence we may regard them as established for all arcs. We will repeat and number the formulas for the purpose of more convenienť reference.

sin a cos b+sin b cos a
sin (a+b)=

sin a cos b--sin b cos a
sin (a=6)

cos a cos b— sin a sin b
cos (a+b)

cos a cos b+sin a sin b
cos (a-


R XX. If, in the formulas of the preceding Article, we make b=a, the first and the third will give 2 sin a cos a

cos? a-sin? a 2 cos” a–R. sin 2a=

cos 2a= R


R formulas which enable us to find the sine and cosine of the double arc, when we know the sine and cosine of the arc itself. To

express the sin d and cos a in terms of ja, put fa for a, and we have 2 sin la cos la

cosa ļa—sin ja R

R To find the sine and cosine of ļa in terms of a, take the equations

cos' a+sin' ?a=R, and cos'ja—sin ja=R cos a, there results by adding and subtracting

cos? ja=R2+R cos a, and sin? ja= RP-R cos a; whence

sin ja=V (RP-R cos a)={V2R?—QR cos a.
cos ja=V (RP+*R cos a)=1V2R2+2R cos a.

sin a=

COS a=.

[ocr errors]

If we put 2a in the place of a, we shall have,

sin a=V (RP-R cos 2a)=V2R%2R cos 2a.

cos a=V (R2 + R cos 2a)=1V 2R”+ 2R cos 2u. Making, in the two last formulas, a=45°, gives cos 2a=0, and

sin 45o=V?R=RV}; and also, cos 45o=V?R=RVI. Next, make a=22° 30', which gives cos 2a=RV}, and we have sin 22° 30'ER V1-V A) and cos 22° 30'=RV (+*V!).

XXI. If we multiply together formulas (1.) and (2.) Art. XIX. and substitute for cosa, R?—sin” a, and for cos” b, R?—sino b; we shall obtain, after reducing and dividing by R?, sin (a+b) sin (ab)=sin’a—sin?b=(sina+sin b) (sina—sin b).

or, sin (ab) : sin a-sin b : : sin a+sin b : sin (a+b).

XXII. The formulas of Art. XIX. furnish a great number of consequences ; among which it will be enough to mention those of most frequent use. By adding and subtracting we obtain the four which follow,

sin (a+b)+sin (ab)= sin a cos b.


2 sin (a+b)—sin (ab) =-.

-sin b cos a.

cos (a+b)+cos (ab)= -cos a cos b.


cos (ab)—cos (a+b)= i sin a sin b.

. and which serve to change a product of several sines or cosines into linear sines or cosines, that is, into sines and cosines multiplied only by constant quantities.

+ XXIII. If in these formulas we put a+b=p, a-b=q, which

p+q P-9 gives a=


we shall find 2


sin p+sin q= sin (p+q) cos }(P9) (1.)


sin p-sin q= sin (P-9) cos } (P+q) (2.)

cos p+cosq= cos }(P+q) cos } (P-9) (3.)


cosq_cosp= sin 1 (P+q) sin } (p9) (4.)


[ocr errors]
[ocr errors]


sin p

: hence

If we make q=0, we shall obtain,

2 sin } p cos IP

& R

2 cos } P R+cos p=


2 sino P R-cos p=


tang P R R+cos P

R cot p
cot } P


R which are often employed in trigonometrical calculations for reducing two terms to a single one.

sin P

sin P

[ocr errors]
[ocr errors]

tang ip:

COS. (l



XXIV. From the first four formulas of Art XXIII. and the first

sin a tang a

R of Art. XX., dividing, and considering that


cot a we derive the following: sin p+-sin

q _sin } (p+q) cos } (p9)_tang 3 (P+q)

sin p-sinq cos. (P+q) sin (p-1) tang 3 (P-9)
sin p+sin q _sin } (p+9) _tang } (p+q)
cos p+cos q cos!


sin p+sin q_cos} (p9)_cot } (P-9)

P-9. cos 9-COS P sin } (P-1) R sin p—sin q _sin } (p9)_tang) (P-1)

p} 9 COS P + cos q cos} (249)

R sin p—sin 9 _cos } (p+9)_cot } (p+9) cos q_COS P sin (P+9) R cos p + cos q _cos? (p+9) cos} (p—-9) _cot } (p+9)

} p+ cos (-COS P sin ) (P+) sin } (P-9) tang(p-1) sin p+sin q 2sin (p+9) cos } (p--9) cos } (p-1) (P+9) 2sin ž (p+q) cosí (p+q) cos ž (p+q)

2 sin p—sin q_2sin (p9) cos } (p+9) sin } (p9)

} } sin (+9)2sin ; (+9) cos (+9) sin (p+9) p į P } P

; P Formulas which are the expression of so many theorems. From the first, it follows that the sum of the sines of two arcs is to the difference of these sines, as the tangent of half the sum of the arcs is to the tangent of half their difference.



XXV. In order likewise to develop some formulas relative to tangents, let us consider the expression

R tang (a+b)=

R sin (a+b), in which by substituting the values

cos (a+b) of sin (a+b) and cos (a+b), we shall find

R (sin a cos b+sin b cos a) tang (a+b) =

cos a cos b-sin b sin a
Now we have sin a=
cos a tang a

cos b tang
and sin b=

R substitute these values, dividing all the terms by cos a cos b; we shall have


(tang a+tang b) tang (a + b)=

R?—tang a tang b which is the value of the tangent of the sum of two arcs, expressed by the tangents of each of these arcs. For the tangent of their difference, we should in like manner find

R? (tang a-tang b)

( -6 Suppose b=a; for the duplication of the arcs, we shall have the formula

2 R2 tang a

R_tang’a Suppose b=2a; for their triplication, we shall have the formula

R” (tang a+tang 2 a) tang 3 a=


R-tang a tang 2 a in which, substituting the value of tang 2 a, we shall have

3R2 tang a—tang 3a tang 3 a=

R3 tang "a.

tang (ab)=R2+tang a tang b:

[ocr errors]

tang 2a

XXVI. Scholium. The radius R being entirely arbitrary, is generally taken equal to 1, in which case it does not appear in the trigonometrical formulas. For example the expression for the tangent of twice an arc when R=1, becomes,

2 tang 1 tang 2 a=.

1-tango a If we have an analytical formula calculated to the radius of 1, and wish to apply it to another circle in which the radius is R, we must multiply each term by such a power of R as will make all the terms homogeneous: that is, so that each shall contain the same number of literal factors.

« PreviousContinue »